即比如我有100個可執(zhí)行文件，互相間沒有特別的先后執(zhí)行關(guān)系，如CODE:
復(fù)制代碼 代碼如下:
job_1
job_2
job_2
.....
job_100
想用csh/bash來多線程調(diào)用執(zhí)行。
比如一次開5個線程，那么job_1,2,3,4,5一起先開始，那么其中任何一個線程如果先執(zhí)行完成，則繼續(xù)執(zhí)行下一個沒有初執(zhí)行過的文件，如job_6,7,8....，這樣一直以所指定的線程數(shù)來執(zhí)行所有100個文件。
我本來想用 "" 來放入后臺，可是這樣我一次可以指定5放入后臺，但是無法知道其中任何一個程序何時執(zhí)行完畢，所以也無法繼續(xù)執(zhí)行下一個程序啊!
完美解決方案：
復(fù)制代碼 代碼如下:
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; cat job_1
#!/bin/bash
n=$((RANDOM % 5 + 1))
echo "$0 sleeping for $n seconds ..."
sleep $n
echo "$0 exiting ..."
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; for ((i = 2; i = 10; ++i)); do cp job_1 job_$i; done
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; cat jobs.sh
#!/bin/bash
nParellel=5
nJobs=10
sJobPattern='./job_%d'
aJobs=()
sNextJob=
for ((iNextJob = 1; iNextJob = nJobs; )); do
for ((iJob = 0; iJob nParellel; ++iJob)); do
if [ $iNextJob -gt $nJobs ]; then
break;
fi
if [ ! "${aJobs[iJob]}" ] || ! kill -0 ${aJobs[iJob]} 2> /dev/null; then
printf -v sNextJob "$sJobPattern" $((iNextJob++))
echo "$sNextJob starting ..."
$sNextJob
aJobs[iJob]=$!
fi
done
sleep .1
done
wait
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; ./jobs.sh
./job_1 starting ...
./job_1 sleeping for 3 seconds ...
./job_2 starting ...
./job_2 sleeping for 2 seconds ...
./job_3 starting ...
./job_3 sleeping for 5 seconds ...
./job_4 starting ...
./job_5 starting ...
./job_4 sleeping for 4 seconds ...
./job_5 sleeping for 2 seconds ...
./job_2 exiting ...
./job_6 starting ...
./job_6 sleeping for 2 seconds ...
./job_5 exiting ...
./job_7 starting ...
./job_7 sleeping for 1 seconds ...
./job_1 exiting ...
./job_8 starting ...
./job_8 sleeping for 3 seconds ...
./job_7 exiting ...
./job_9 starting ...
./job_9 sleeping for 5 seconds ...
./job_4 exiting ...
./job_6 exiting ...
./job_10 starting ...
./job_10 sleeping for 5 seconds ...
./job_3 exiting ...
./job_8 exiting ...
./job_9 exiting ...
./job_10 exiting ...
-(dearvoid@LinuxEden:Forum)-(~/tmp)-
[$$=6718 $?=0] ; bye
